如何使一个Python类序列化?
class FileItem:
def __init__(self, fname):
self.fname = fname
尝试序列化为JSON:
>>> import json
>>> x = FileItem('/foo/bar')
>>> json.dumps(x)
TypeError: Object of type 'FileItem' is not JSON serializable
如何使一个Python类序列化?
class FileItem:
def __init__(self, fname):
self.fname = fname
尝试序列化为JSON:
>>> import json
>>> x = FileItem('/foo/bar')
>>> json.dumps(x)
TypeError: Object of type 'FileItem' is not JSON serializable
当前回答
import simplejson
class User(object):
def __init__(self, name, mail):
self.name = name
self.mail = mail
def _asdict(self):
return self.__dict__
print(simplejson.dumps(User('alice', 'alice@mail.com')))
如果使用标准json,则需要定义一个默认函数
import json
def default(o):
return o._asdict()
print(json.dumps(User('alice', 'alice@mail.com'), default=default))
其他回答
这个类可以做到这一点,它将object转换为标准json。
import json
class Serializer(object):
@staticmethod
def serialize(object):
return json.dumps(object, default=lambda o: o.__dict__.values()[0])
用法:
Serializer.serialize(my_object)
在python2.7和python3中工作。
一个非常简单的一行程序解决方案
import json
json.dumps(your_object, default=lambda __o: __o.__dict__)
结束!
下面是一个测试。
import json
from dataclasses import dataclass
@dataclass
class Company:
id: int
name: str
@dataclass
class User:
id: int
name: str
email: str
company: Company
company = Company(id=1, name="Example Ltd")
user = User(id=1, name="John Doe", email="john@doe.net", company=company)
json.dumps(user, default=lambda __o: __o.__dict__)
输出:
{
"id": 1,
"name": "John Doe",
"email": "john@doe.net",
"company": {
"id": 1,
"name": "Example Ltd"
}
}
import simplejson
class User(object):
def __init__(self, name, mail):
self.name = name
self.mail = mail
def _asdict(self):
return self.__dict__
print(simplejson.dumps(User('alice', 'alice@mail.com')))
如果使用标准json,则需要定义一个默认函数
import json
def default(o):
return o._asdict()
print(json.dumps(User('alice', 'alice@mail.com'), default=default))
下面是一个简单功能的简单解决方案:
.toJSON()方法
实现一个序列化器方法,而不是一个JSON可序列化类:
import json
class Object:
def toJSON(self):
return json.dumps(self, default=lambda o: o.__dict__,
sort_keys=True, indent=4)
所以你只需调用它来序列化:
me = Object()
me.name = "Onur"
me.age = 35
me.dog = Object()
me.dog.name = "Apollo"
print(me.toJSON())
将输出:
{
"age": 35,
"dog": {
"name": "Apollo"
},
"name": "Onur"
}
前几天我遇到了这个问题,并为Python对象实现了一个更通用的Encoder版本,可以处理嵌套对象和继承字段:
import json
import inspect
class ObjectEncoder(json.JSONEncoder):
def default(self, obj):
if hasattr(obj, "to_json"):
return self.default(obj.to_json())
elif hasattr(obj, "__dict__"):
d = dict(
(key, value)
for key, value in inspect.getmembers(obj)
if not key.startswith("__")
and not inspect.isabstract(value)
and not inspect.isbuiltin(value)
and not inspect.isfunction(value)
and not inspect.isgenerator(value)
and not inspect.isgeneratorfunction(value)
and not inspect.ismethod(value)
and not inspect.ismethoddescriptor(value)
and not inspect.isroutine(value)
)
return self.default(d)
return obj
例子:
class C(object):
c = "NO"
def to_json(self):
return {"c": "YES"}
class B(object):
b = "B"
i = "I"
def __init__(self, y):
self.y = y
def f(self):
print "f"
class A(B):
a = "A"
def __init__(self):
self.b = [{"ab": B("y")}]
self.c = C()
print json.dumps(A(), cls=ObjectEncoder, indent=2, sort_keys=True)
结果:
{
"a": "A",
"b": [
{
"ab": {
"b": "B",
"i": "I",
"y": "y"
}
}
],
"c": {
"c": "YES"
},
"i": "I"
}